- #1
Elvis 123456789
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Homework Statement
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 23.5m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 14.8m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.20s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 23.5m/s.
How high was the balloon when the rock was thrown out?
Homework Equations
ΔY=V1y*t + 1/2*a*t^2
The Attempt at a Solution
At first I thought since the stone is being thrown off the balloon which is going at a constant speed of 23.5m/s, then the stone's initial Y velocity must also be 23.5m/s, and since its being thrown perpendicular to the balloon's path at 14.8m/s, then that must be its initial X velocity.
I figure that the balloon's height when the stone was thrown out, is the same as the stone's initial position. And the stone's final position is 0 at t=6.20 s since its on the ground.
So I used the above equation
ΔY=V1y*t + 1/2*a*t^2
0-y1=V1y*t - 1/2*g*t^2
y1= -[(23.5m/s)(6.20s)-(4.9m/s^2)(6.20s)^2
y1=47.356m
but this isn't right and I don't know why